21 Mar 2019

# Visualize $$e^{i\theta}$$

$$e^{i\theta}$$ is the points on unit circle, at angle $$\theta$$, here is a visualization of how such circle is derived from the sum of power series:

Assume we have a function with the property of its derivative is equal to itself, just as $$\frac{d}{dx}e^x = e^x$$, such function should have the form:

$f(x) = e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$

Expand to complex number $$i\theta$$:

$e^{i\theta} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + ...$

Draw these complex numbers as vectors on complex plane, we have a end point for each $$\theta$$, and a unit circle with all $$\theta$$s

move mouse horizontally to draw the circle

# Complex number as a Power Series

Any point on C plane can be written as

$a = x + iy = e^z = 1 + z + \frac{1}{2!}z^2 + \frac{1}{3!}z^3 + ...$

Let’s first compute the value of z using complex logarithm:

$z = log(a) = log(\vert a \vert) + i\arctan(a)$

each term in the series is a vector, we add up these vectors to recover a in complex plane.

move mouse to draw the vectors

# Euler’s formula

$$e^{iy} = \cos y + i \sin y$$ corresponds to imagenary axis

# Compound interest

$$e^{z} = e^{x+iy} = e^x e^{iy} = (1 + \frac{z}{n})^n$$ if n is large

# Inverse

$z = r e^{i\theta}$ $\frac{1}{z} = \frac{1}{r} e^{-i\theta}$

The inverse can be seen the image of z wrt. the unit circle, the mapping has some interesting trajectories, draw some strokes below to explore…

move mouse to draw 1/z
function drawETheta(theta, iter, scale) {
var itheta = new Complex(0, theta);
var res = new Complex(1, 0).add(itheta);

var path = new Path();
path.strokeColor = 'black';

var itheta_pow = new Complex(itheta);

for (var i = 2; i < iter; i++) {
itheta_pow = itheta_pow.multiply(itheta);
}
return path;
}

function drawExpZ(z, iter) {
// res = 1 + z
var w = z.log();
var res = new Complex(1 * scale, 0).add(w * scale);
var w_pow = new Complex(w);

var path = new Path();
path.strokeColor = 'black';